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I heard today that deuterium for nuclear fusion could be extracted from the Oceans, in which it is present in "heavy water." I heard this claim:

"Deuterium is distributed uniformly with Ocean depth."

This contradicted my naive expectation that deuterium density would increase with depth, on account of it being in "heavy water." For example, salinity increases with depth. I asked, and received two contradictory explanations.

The first was that because powerful waves and Ocean currents mix Ocean water, gravity is negligible. This surprised me; I thought that Ocean's were stratified, with predominantly tangential, rather than vertical, currents.

The second explanation was that my view of seawater is naive - that I should think of seawater not as heavy and regular water molecules, but as a soup of weakly interacting deuterium, the normal hydrogen isotope and oxygen, forming and breaking water and heavy water molecules with a short time-scale. If so, apparently, deuterium density and the normal hydrogen isotope would be distributed evenly with depth. This did not help me, because, I then naturally wondered why deuterium's density doesn't increase with depth, since it is heavier than hydrogen-1?

Is the claim correct? and if so, what is really going on? I have the feeling that Ocean behaviour with depth might be non-trivial...

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A little thermodynamics, with the tradeoff between entropy and potential energy, would go a long way here. Also, there are definitely deep ocean currents doing all sorts of cool things slowly but steadily. Hopefully an oceanographer steps in to make a better answer; otherwise I'll be forced to figure this out from an astrophysics point of view. – Chris White Jul 3 '13 at 22:48
up vote 5 down vote accepted

Reading a discussion on this topic on the XKCD forums, it seems there is a whole book dedicated to this question. If you check e.g. page 51, the graph there indicates a much more detailed structure than an uniform one.

To be more precise, there are several mechanisms in play:

  • evaporation
  • ice formation
  • ice melting
  • influx of fresh water

Several quotes:

Of all natural waters, the ocean, which is a unique reservoir, remains most constant regarding its isotopic and other physicochemical properties. Several authors (Craig and Gordon 1965; Craig 1961b; Epstein and Mayeda 1953; Friedman et al. 1964) have shown that ocean water at depth of more than 500 m is homo- geneous in isotopic composition. This allowed Craig (1961b) to propose it as a standard for reporting concentrations of $D$ and $\phantom0^{18} O$ content in natural water.

The variation in the relative content of $D$ in deep ocean layer is about 4 per mil and that of $\phantom0^{18} O$ is of 0.3 per mil. In the surface ocean layer, the regional variations, depending upon water temperature, are 35 per mil for $D$ and about 3 per mil for $\phantom0^{18}O$. The lowered content of $D$ in the surface ocean layer occurs in those regions where ice-melting water affects isotopic composition. In high latitudes, where the surface layer of the ocean is freezing, isotopic fractionation factor is abt. 1.0180 for $D$ and abt. 1.0030 for $\phantom0^{18} O$. In the equatorial region of the ocean where intensive evaporation of water takes place, there is an enrichment of the surface layer with heavy isotopes.

Despite the considerable homogeneity of the isotopic composition of ocean water, the ranges of $D$ and $\phantom0^{18} O$ variations with latitude and depth are wide enough to be reliable indicators of processes occurring in the ocean. These ranges, for deep ocean waters, are about ten times greater than the accuracy of mass spectrometer measurements with which modern techniques and apparatus permit determination of $D$ (+-0.2 per mil) and $\phantom0^{18}O$ (+-0.02 per mil).

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Thanks, but, lazy though it might sound, I don't want to read a whole book on Oceanography. I was hoping someone with expertise could precis the relevant physics to give me an idea of what's going on. – innisfree Jul 3 '13 at 22:43
@innisfree - if you don't want to read a page, there's nothing else we can do for you. Wojciech found the right book (V. Ferronsky, V. Polyakov. Isotopes of the Earth's Hydrosphere, 2012) and it contains all the details on fractionation due to evaporation and ice formation. – Deer Hunter Jul 4 '13 at 5:31
@DeerHunter I am happy to read a page, but don't have time for the whole book. The page in question, in isolation, wasn't that helpful for me. Besides, what is this site for other than to summarize physics for curious people who don't have time to read whole books? Other than research questions, most questions here could be directed to text books. I thought it was site policy that answers should be self-contained and not simply e.g. weblinks. – innisfree Jul 4 '13 at 8:48
@innisfree - I have already edited Wojciech's post; the edit is currently pending review, and you can ping a mod to speed up the process :) – Deer Hunter Jul 4 '13 at 9:36

Previous answers make it clear that there are many factors that could lead to temporary non-uniform distributions of deuterium atoms in the majority of "normal" hydrogen atoms in sea-water. So, if one were to pour a beaker of $D_2O$ carefully and slowly into a beaker of normal water, the heavy water would be found concentrated at the bottom of the beaker.

However, the OP's question seems to me to be: If we mix the stratified beaker above, will the higher concentration of deuterium at the bottom of the beaker be restored?

In my opinion, the answer is yes, with a serious reservation. The concentration would depend on how large the difference in gravitational potential energy is, compared to the basic kinetic energy of the atoms in the solution. My thought is: not much. How often do you find a layer of alcohol at the top of an long-undisturbed bottle of beer?

In a similar vein, leaking chlorine (from, say a tank car) does concentrate in lower areas. However, a gymnasium full of thoroughly mixed chlorine-contaminated air will never have a centimeters-thick layer of poison at floor level. The gas molecules are travelling at speeds of the order of the speed of sound. Just compare $\frac{1}{2}v^2$ with $gh$.

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